07 Free Fall (Distance).pptx

We now want to learn how

FAR it travels in any TIME.

When an object

ACCELERATES UNIFORMLY,

we know how FAST

it is traveling at any TIME.

I can . . .

• _{apply velocity and acceleration to}

determine the distance an object moves as a function of time.

• apply the appropriate kinematic equation to solve a problem.

• explain the meaning of the area under the curve of a graph involving position,

velocity, or acceleration with respect to

time.

http://static.howstuffworks.com/gif/speedometer-1.jpg 0 10 20 30 40

50 60 70 80 90

100 110

120 130 meters per second

The velocity of an object in freefall changes by

approximately

10 m/s each second.

An object in freefall accelerates at

approximately 10 m/s2

0 10 20 30 40 50 60 70 80 90 100 110 120 130

0 1 2 3 4 5 6 7 8 9 10 11 12 13

http://static.howstuffworks.com/gif/speedometer-1.jpg 0 10 20 30 40

50 60 70 80 90

100 110

120 130 meters per second

The velocity of an object in freefall changes by

approximately

10 m/s each second.

An object in freefall accelerates at

approximately 10 m/s2

0 10 20 30 40 50 60 70 80 90 100 110 120 130

0 1 2 3 4 5 6 7 8 9 10 11 12 13

If and object went 10 m/s for the whole first second, it would have traveled 10 meters in that second.

The object actually started at zero m/s and accelerated up to 10 m/s,

but it only AVERAGED 5 m/s.

It only traveled 5 meters during the first second.

10 9 8 7 6 5 4 3 2 1 0 0+1+2+3+4+5+6+7+8+9+10 11

d = · 1sec

d = 10 m/s · 1 s

d = v · t

d = 10 meters

d = 5 meters

m/s m/s m/s

s s s s s

0

10

20

30

40

50

60

70

80

90

100

110

120

130

0

1

2

3

4

5

6

7

8

9 10 11 12 13

An object in free fall travels FARTHER each second because it is moving FASTER each second.

Free Fall

0 1 2 3 4 5 6 7 8 9 10

Average Velocity =

Distance

_Time

V =

d

_t

d = V · t

How far will you travel if you

AVERAGE

50 m/s

for

3 s

?

V =

V

f

+ V

0

2

Average Velocity for

Constant Acceleration

What was your

average

velocity

if you accelerate

from

20 m/s

to

30 m/s

?

V =

V

+ V

2

Distance Traveled for:

Time V e lo ci ty Constant VELOCITY

d = V · t

Time V e lo ci ty Uniform ACCELERATION

d = · t

V

+ V

If dropped from rest,

V

= 0.

V =

V

f

2

Average Velocity for

Constant Acceleration

What was your

average velocity

if you

accelerate from

REST

to

30 m/s

?

15 m/s

0 10

20 30

40 50 60 70 8090 100

110 120 130 meters per second

Acceleration = 3 m/s

?

Time = seconds

10

012345678

9

How far did the

object travel

during the

10 seconds?

d = V · t

0 10

20 30

40 50 60 70 8090 100

110 120 130 meters per second

Acceleration = 4 m/s

?

Time = seconds

012345

How far did the

object travel

during the

5 seconds?

d = V · t

Time (sec) Instantaneou s Velocity (m/s) Distance Traveled EACH second (m) TOTAL Distance Traveled (m) 0 0 1 2 3 4 5

Car Accelerates a = 8 m/s

8 m/s 16 m/s 24 m/s 32 m/s 40 m/s 4 m 12 m 20 m 28 m 36 m 4 m 16 m 36 m 64 m 100 m

0 s 1 s 2 s 3 s 4 s 5 s

0 m/s 8 m/s 16 m/s 24 m/s 32 m/s 40 m/s

Time (sec) Instantaneou s Velocity (m/s) Distance Traveled EACH second (m) TOTAL Distance Traveled (m) 0 0 1 2 3 4 5

Ball is dropped g ≈ 10 m/s

30 m/s 20 m/s 10 m/s 0 m/s 40 m/s 80 m 45 m 20 m

5 m _{3 sec}

2 sec 1 sec 0 sec 4 sec 1 sec 2 sec 3 sec 4 sec If you throw a ball up at 40 m/s,

How long will it be in the air? How high will it go?

If a ball is thrown so it is in the air for 6 seconds,

How fast was it thrown? How high will it go?

8 seconds 80 meters

a·t

d = V · t

_{V =}

V

2

d = · t

V

_f

2

d = · t

₂

d = ½ a · t

2

d = ½·a·t

2

d = ½·g·t

2

(Any Acceleration)

(Acceleration of GRAVITY)

Time (sec) Instantaneous Velocity (m/s) Distance Traveled EACH second (m) TOTAL Distance Traveled (m)

0

0

1

10 m/s

5 m

5 m

2

20 m/s 15 m

20 m

3

30 m/s 25 m

45 m

4

40 m/s 35 m

80 m

5

50 m/s 45 m 125 m

g = 10 m/s

d = ½ a · t

d = ½ 10 · 1

d = ½ 10 · 2

d = ½ 10 · 3

d = ½ 10 · 4

www.nolimitstahoe.com/ adventures/photos.htm

Meters

d = ½(9.80)·t

2

Feet

d = ½(32.2)·t

2

El Capitan is 3,000 ft tall

3,000 = ½(32.2)·t

2

= 14 s

1

.

16

3000



8.0 seconds

Feet d = ½(a)·t

2

d = ½(32.2)·8.0

2

Lyons Ferry Bridge

165 = ½(32.2)·t

2

= 3.20 s

Feet

d = ½·a·t

2

V

= a · t

V

= 32.2 · 3.20

V

= 103

ft

_s

_{· ·}

_{5280 ft}

1 mi

3600 s

_{1 hr}

V

= 70.3

mi

_hr

1

.

16

165



seconds

0123456789

V

= a · t

V

= 290

ft

_s

_{· ·}

_{5280 ft}

1 mi

3600 s

_{1 hr}

V

= 200

mi

_hr

Feet

d = ½·a·t

2

d = ½(32.2)·9.0

2

d = 1300 ft

V

= 32.2 · 9.0

_ss

ft

s

m

m

s

s

+

m

s

s

x

y

70.0 m

x

y

-70.0 = 0 + 12.0·t + 1/2(-9.8)t

-70.0

0 10

20 30

40 50 60 70 8090 100

110 120 130

Time = seconds

10

012345678

9

V

₀

= 50 m/s

Hang Time

Time = seconds

10

V

₀

= 50 m/s

Hang Time

http://upload.wikimedia.org/wikipedia/commons/d/d3/Spiegel_Building_Hamburg_3.jpg

How high did the ball go?

d = ½·a·t

d = ½·10·

5

I can . . .

• _{apply velocity and acceleration to}

determine the distance an object moves as a function of time.

• apply the appropriate kinematic equation to solve a problem.

• _{explain the meaning of the area under the}

curve of a graph involving position,

velocity, or acceleration with respect to

time.

d = V·t

d = 50 m/s · 5 s

d = 250 m

x = ½·a·t

x = ½·10·5

x = 125 m

x = 25 m/s · 5 s

x = 125 m

x = ½ 50 m/s · 5 s

x = V·t

x = v

·t + ½·a·t

x = 10·5 + ½·10·5

x = 175 m

x = ½ (10+60) · 5

x = 175 m

x = 35m/s · 5 s

x = V·t

Area under the “Curve”

x = A

+ A

x =

10·5

+

½·50·5

20·2

½(60+20)·2 ½(20+30)·1

40 m

80 m 25 m

145 m

½ (-20)·2 ½(30)·3

-20 m 45 m

25 m

What does the area under

the curve give you?

A

cc

el

er

at

io

n

(

m

/s

The moment the sports car traveling at 20 m/s

passes the police car at rest, the police car begins to

accelerate at a rate of 4 m/s2.

Find the time when the police car catches up to the

sports car:

The moment the sports car traveling at 20 m/s passes the police car

Acceleration is CONSTANT.

Velocity is INCREASING each second

1 second

10 m/s

1 second

10 m/s

Area = Base · Height m

s2

= s ·

= m/s (velocity)

Slope = Rise Run Δ Y Δ X = m/s s = m s2 =

Area = ½ Base · Height m

s = s ·

= m (distance)

Slope = Rise Run Δ Y Δ X = m s =

= (velocity)

2 seconds

40 meters 2 seconds

Slope = = = .40 m/ s

13 – (-1) 45 - 10

14 35

(50,13)

Slope = = .26 m/s 13 50

(50, 13)

Slope = = .32 m/s 13 –(-3) 50 - 0

y

=

5

·

x

+

0

y

=

m

·

x

+

b

d

= (

½g

)(

t

)